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The coordinates of the centre of a circle passing through (1, 2), (3, – 4) and (5, – 6) is:
One end of a line of length 10 units is at the point (3, 2). If the ordinate of the other end be 10, then the abscissa will be
The horizontal and vertical lines drawn to determine the position of a point in a Cartesian plane are called
The point is determined on 2 dimensions .The name of horizontal and vertical lines drawn to determine the position of any point in the Cartesian plane is xaxis and yaxis respectively. The name of each part of the plane formed by these two lines xaxis and the yaxis is quadrants. The point where these two lines intersect is called the origin.
Let P(x, y) be equidistant from the points A (7, 1) and (3, 5).Find a relation between x and y.
If the four points (0,1), (6,7),(2,3) and (8,3) are the vertices of a rectangle, then its area is
Let A(01), B(6,7), C(2,3) and D(8,3) be the given points. Then
∴ AD = BC and AC = BD
So, ADBC is a parallelogram
Now
Clearly, AB^{2 }= AD^{2 }+ DB^{2 }and CD^{2 }= CB^{2 }+ BD^{2}
Hence, ADBC is a rectangle.
Now
,Area of rectangle ADBC = AD × DB =(4√5 × 2√5)sq. units = 40sq. units
The perimeter of a triangle with vertices (0, 4) (0, 0) and (3, 0) is:
The ordinate of a point is twice its abscissa. If its distance from the point (4,3) is √10, then the coordinates of the point are
The perimeter of the triangle formed by the points A(0,0), B(1,0) and C(0,1) is
The value of k, if the point P(0,2) is equidistant from A(3,k) and B(k,5) is
If A and B are the points (6, 7) and (1, 5) respectively, then the distance 2AB is equal to
The values of x and y, if the distance of the point (x,y) from (3,0) as well as from (3,0) is 4 are
Distance formula=
The point on yaxis that is equidistant from (2,3) and (4,1) is
A(2, 3) and B(4, 1) are the given points.
Let C(0.y) be the points are y  axis
Therefore, The point on y  axis is (0,1)
The point on xaxis which is equidistant from (5,9) and (4,6) is
The distance of the point P(6,6) from the origin is equal to
The distance between the points P (6,7) and Q (1,5) is
The condition that the point (x,y) may lie on the line joining (3,4) and (5,6) is
Since the point P(x,y) lies on the line joining A(3,4) and B(5,6),
Therefore, points P, A and B are collinear points.
So, area of triangle PAB = 0
Therefore, we have:
10x183y5y+20=0
10x8y+2=0
5x4y+1=0 , which is the required condition.
The points (3, 2), (0, 5), (3, 2) and (0, 1) are the vertices of a quadrilateral. Which quadrilateral is it?
x=3 is fixed. This means the value of x is constant. So y can vary but x has only one value. For example (3,0),(3,2),(3,5) etc. So the line drawn will be parallel to y axis as y can vary.
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